
#1
Jan1513, 03:40 PM

P: 73

Following is a list of 39 Mersenne numbers. Some are prime and some are not. These are generated with x =( 2^n) 1. Many of the largest primes known are Mersenne primes.
I would like to point out that a sieve may be used to block out many non prime Mersenne numbers. For example for n = 2 x = 3 but on closer inspection every even value of n other than n = 2 is non prime furthermore they are all divisible by 3, Thus we can cross out n=4, n=6, … up to n=38. Also starting at n=3 and x = 7 we find that every third number larger than 3 is divisible by 7 so we could cross out n = 6, n=9, n=12 … up to n = 39. Now lets move to n=5 or x = 31. now we can cross out every fifth number larger than n=5 as they are all divisible by 31 if we move to n=7 and x = 127 we find that every 7th number larger than 7 is divisible by 127 and can be crossed out. 11 should be crossed off as 2047 is not prime My question is does this go on forever and can we use it to delete many Mersenne numbers as non primes. The next prime is 11 n 2^n  1 n = 1 x = 1 n = 2 x = 3 n = 3 x = 7 n = 4 x = 15 n = 5 x = 31 n = 6 x = 63 n = 7 x = 127 n = 8 x = 255 n = 9 x = 511 n = 10 x = 1023 n = 11 x = 2047 n = 12 x = 4095 n = 13 x = 8191 n = 14 x = 16383 n = 15 x = 32767 n = 16 x = 65535 n = 17 x = 131071 n = 18 x = 262143 n = 19 x = 524287 n = 20 x = 1048575 n = 21 x = 2097151 n = 22 x = 4194303 n = 23 x = 8388607 n = 24 x = 16777215 n = 25 x = 33554431 n = 26 x = 67108863 n = 27 x = 134217727 n = 28 x = 268435455 n = 29 x = 536870911 n = 30 x = 1073741823 n = 31 x = 2147483647 n = 32 x = 4294967295 n = 33 x = 8589934591 n = 34 x = 17179869183 n = 35 x = 34359738367 n = 36 x = 68719476735 n = 37 x = 137438953471 n = 38 x = 274877906943 n = 39 x = 549755813887 



#2
Jan1513, 11:28 PM

P: 96

You should look at: http://en.wikipedia.org/wiki/Mersenne_primes
There you can see, that n in your notatation must be prime to let 2**n  1 be prime 



#3
Jan1613, 04:45 AM

P: 96

As a consequence of Theorem 18 from HardyWright, we have the following
Corollary: For two natural numbers 1 < a and b: a[itex]{^b}[/itex]  1 is composite if a > 2 (because (a  1) divides a[itex]^{b}[/itex]  1); or in the case a = 2: if b = s * t (because 2[itex]^{s}[/itex]  1 divides 2[itex]^{s*t}[/itex]  1 



#4
Jan1613, 02:31 PM

P: 73

Mersenne Sieve
yes but consider the number 11 which is a prime while 2^11 1 is not a prime.




#5
Jan1613, 02:33 PM

P: 73

yes but this does not help




#6
Jan1713, 03:03 AM

P: 96

I think, my Corollary helps to delete all 2[itex]{^b}[/itex]  1 with a composite b; but unfortunately, there is no help for sieving the Mersenne primes:
Theorem 18 by HW says (in short): 'If 2[itex]{^b}  1[/itex] is a prime, then b is a prime'; and the 'other way round' is not valid 



#7
Jan1713, 08:06 AM

P: 73

To show that (X^(2*n) )  1 has a factor of x1 for any n.
Consider x^2  1 = (x1) * (x+1 ) so if x^2 1 is a factor of a function then x1 is a factor. Consider x^4  1 = ( x^2 1 ) * ( x^2 + 1 ) so x1 is a factor of x^4 1 consider x^6  1 = (x^2 1 ) * ( x^4 + x^2 + 1 ) so x1 is a factor of x^6 1 consider x^81 = ( x^2 1 ) * ( x^6 + x^ 4 + x^ 2 + 1 ) so x1 is a factor of x^8 1 etc So it seems to me… x^(2*n)  1 = (x^2 1 ) * ( x^(2*(n1)) + x^(2 * (n2)) + ……. + 1 ) 


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